# (solved)Question 1.34 of NCERT Class XI Chemistry Chapter 1

A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate (i) empirical formula, (ii) molar mass of the gas, and (iii) molecular formula.
(Rev. 16-May-2023)

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### Question 1.34 NCERT Class XI Chemistry

A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate (i) empirical formula, (ii) molar mass of the gas, and (iii) molecular formula.

### Video Explanation(detailed solution given after this video)

Please watch this youtube video for a quick explanation of the solution:

### Solution in Detail

Let the empirical formula be $\displaystyle C_xH_y$

$\displaystyle C_xH_y + \text{oxygen} \rightarrow \underset{3.38 \text{ g}}{\text{CO}_2} + \underset{0.690 \text{ g}}{\text{H}_2\text{O}}$

$\displaystyle \implies \text{(1) C in C}_x\text{H}_y = \text{C in CO}_2$

$\displaystyle \implies \text{(2) H in C}_x\text{H}_y= \text{H in H}_2\text{O}$

$\displaystyle \underline{\underline{\text{Step 1: Calculate C in 3.38 g CO}_2}}$

Molar mass of $\displaystyle \text{CO}_2 = 44 \text{ g/mol}$

mols of $\displaystyle \text{CO}_2 = \frac{3.38}{44} = \frac{169}{2200}$

i.e., $\displaystyle \text{mols of C } \approx \frac{7.68}{100}\text{. . . (1)}$

$\displaystyle \underline{\underline{\text{Step 2: Calculate H in 0.690 g H}_2\text{O}}}$

Molar mass of $\displaystyle \text{H}_2\text{O} = 18 \text{ g/mol}$

mols of $\displaystyle \text{H}_2\text{O} = \frac{0.690}{18} = \frac{23}{600}$

$\displaystyle \therefore \text{mols of H} = 2 \times \bigg(\frac{23}{600}\bigg)$

$\displaystyle \implies \text{mols of H} = \bigg(\frac{23}{300}\bigg)$

i.e., $\displaystyle \text{mols of H } \approx \frac{7.66}{100}\text{. . . (2)}$

$\displaystyle \underline{\underline{\text{Step 3: Empirical Formula}}}$

by (1) and (2), simplest ratio of C : H is 1 : 1

Hence empirical formula is $\displaystyle CH\:\underline{ Ans}$

$\displaystyle \underline{\underline{\text{(ii) Molecular Mass}}}$

Given 10 L at STP weighs = 11.6 grams

so 22.4 L = $\displaystyle \frac{11.6}{10} \times 22.4 \approx 26 \text{ gms}$

But 1 mol gas at STP occupies 22.4 L

Hence 26 grams is the mass of 1 mol

hence molar mass $\displaystyle = 26\text{ g/mol} \:\underline{ Ans}$

$\displaystyle \underline{\underline{\text{(iii) Molecular Formula}}}$

Empirical formula $\displaystyle CH$

$\displaystyle \therefore$ molecular formula $\displaystyle C_nH_n$

$\displaystyle \therefore$ molar mass = $\displaystyle 12n + 1n = 13n$

$\displaystyle \implies 13n = 26$ (g/mol) found above

$\displaystyle \therefore n = 2$

$\displaystyle \implies$ molecular formula is $\displaystyle C_2H_2 \:\underline{Ans}$